Quantitative analysis of compounds is very important in organic chemistry. It helps chemists in the determination of mass per cent of elements present in a compound. You have learnt in Unit-1 that mass per cent of elements is required for the determination of emperical and molecular formula.
The percentage composition of elements present in an organic compound is determined by the following methods:
12.10.1 Carbon and Hydrogen
Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.
CxHy + (x + y/4) O2 → x CO2 + (y /2)H2O
The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Figure).
The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated. Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m1 and m2 g respectively;
Percentage of carbon = (12 X m2 X 100) / (44 X m)
Percentage of hydrogen = (12 X m1 X 100) / (18 X m)
There are two methods for estimation of nitrogen:
(i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method:
The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
CxHyNz + (2x + y/2)CuO → xCO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu
Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Figure).
Let the mass of organic compound = m g
Volume of nitrogen collected = V1mL
Room temperature = T1K
Volume of nitrogen at STP = (P1V1 x 273 )/ (760 X T1)
Where p1 and V1 are the pressure and volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1 is obtained by the relation; p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g.
V mL N2 at STP weighs = (28 x V) / 22400
Percentage of nitrogen = (28 x V x 100)/(22400 X m)
(ii) Kjeldahl’s method
The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Figure).
The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction.
It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.
Organic compound + H2SO4 → (NH4)2SO4 → Na2SO4 + 2NH3 + 2H2O → 2NH3 + H2SO4 (NH4)2SO4
Let the mass of organic compound taken = m g
Volume of H2SO4 of molarity, M,
taken = V mL
Volume of NaOH of molarity, M, used for titration of excess of H2SO4 = V1 mL
V1mL of NaOH of molarity M = V1/2 mL of H2SO4 of molarity M Volume of H2SO4 of molarity M unused
= (V - V1/2)mL
(V - V1/2)mL of H2SO4 of molarity M = 2(V - V1/2) mL of NH3 solution contains 17g NH3 or 14g of N
2(V – V/2)mL of NH3 olution of molarity M contains:
[(14 x M x 2(V -V1/2)/1000 ]g N
percentage of N = [(14 x M x 2(V -V1/2)/1000] X [100/m] = [1.4 x M x 2 (V -V1/2)]/m
Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.
Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, in a furnace.
Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Let the mass of organic compound taken = m g
Mass of AgX formed = m1 g
1 mol of AgX contains 1 mol of X
Mass of halogen in m1 g of AgX = atomic mass of X x m1 g / molecualr mass of AgX
Percentage of halogen = atomic mass of X x m1 x 100 / molecualr mass of AgX x m
A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.
Let the mass of organic ocmpound taken = m g
and the mass of barium sulphate formed = m1 g
1 mol of BaSO4 = 233 g BaSO4 = 32g sulphur
m1g of BaSO4 contains [(32 x m1)/233] g of Sulphur
Percentage of Sulphur = (32 x m1 x 100) / (233 x m)
A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and ammonium molybdate.
Alternatively, phosphoric acid may be precipitated as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7. Let the mass of organic compound taken = m g and mass of ammonium phospho molydate = m1g
Molar mass of (NH4)3PO4.12MoO3 = 1877 g
Percentage of phosphorus = (31 x m1 x 100)/(1877 x m) %
If phosphorus is estimated as Mg2P2O7
Percentage of phosphorus = (62 x m1 x 100)/(222 x m) %
where, 222 u is the molar mass of Mg2P2O7, m, the mass of organic compound taken, m1, the mass of Mg2P2O7 formed and 62, the mass of two phosphorus atoms present in the compound Mg2P2O7.
The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide.
This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine.
compound → O2 + other gaseous products
2C + O2 → 2CO x 5
I2O5 + 5CO → I2 +5CO2 x 2
On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively;
we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.
Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated. Let the mass of organic compound taken be m g Mass of carbon dioxide produced be m1 g
m1 g carbon dioxide is obtained from [(32 x m1)/88] g of O2
Percentage of oxygen = [(32 x m1 x 100)/88 x m ] %
The percentage of oxygen can be derived from the amount of iodine produced also. Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques.
The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time. A detailed discussion of such methods is beyond the scope of this book.